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mathcalculus
related rate problem: help please. (attached)
Recall the distance formula: \[d^2=(x-x_0)^2+(y-y_0)^2\] In this particular problem, you have \((x_0,y_0)=(0,0)\) (the origin), since you want to find the distance between any given point on the curve \((x,y)\) and the origin. \[d^2=x^2+y^2\] Substitute the given equation of the curve: \[d^2=x^2+\left(4\sqrt{2x+2}\right)^2\\ d^2=x^2+16(2x+2)\\ d^2=x^2+32x+32\] Now differentiate both sides implicitly with respect to some dummy variable \(t\): \[2d\frac{dd}{dt}=(2x+32)\frac{dx}{dt}\] You're told that at the point \((1,8)\), \(x\) is increasing at a rate of 3 units/s, which translates to \(\dfrac{dx}{dt}=3\). \[2d\frac{dd}{dt}=(2\cdot1+32)(3)\\ 2d\frac{dd}{dt}=102\] You're asked to find \(\dfrac{dd}{dt}\), so first you need to find \(d\). To do this, find the distance between \((1,8)\) and \((0,0)\), then plug it into the above equation. Lastly, solve for \(\dfrac{dd}{dt}\).
@SithsAndGiggles omg thank you!! :)
@SithsAndGiggles hey, i found the distance which is 8 right? then i plugged in 8 into the equation and got 51/8... however, something is wrong?
i get \(\frac{19}{8}\)
to find the distance is this:
i got it... thank you! @satellite73
\[d^2=x^2+32x+32\] \[2dd'=2xx'+32x'\] \[dd'=xx'+32x'\] \[8d'=3+96\]
distance isn't 8 is it? distance is \(\sqrt{65}\)
which is pretty close to 8...
is the final answer \(\frac{99}{\sqrt{65}}\) ?
close yes... i thought i had to do this: x2-x1/y2-y1
it's 51/sqrt(65)/65
oh damn i forgot to divide 32 by 2 must be past my bedtime